\chapter{Sheaves}
\section{Sheaves for a Grothendieck Topology}
\begin{Ex}[Disproving the example circulated on 06 Oct 2012.]
\end{Ex}
\begin{proof}[Disproof]
Working on almost the same settings of the mentioned example, Let $\clC$ be the category of Zariski open subset of $X=\PR^2_{\KK}$ and the morphisms of inclusions, for some algebraically closed field $\KK$. And let $\clF$ be ANY functor $\clC^{op}\rightarrow \clA$, where $\clA$ is ANY category with all products.\\
The considered covering is $\{f_i:U_i\rightarrow X,f=1..3\}$. then we will have the diagrams:
$\xymatrix{
U_1\ar[r]^{id_{U_1}}\ar[d]_{id_{U_1}} &U_1 \ar[d]^{f_1}\\
U_1\ar[r]_{f_1}&X
}
\xymatrix{
U_2\bigcap U_1\ar[r]^{\iota_{2,1,1}}\ar[d]_{\iota_{2,1,2}} &U_1 \ar[d]^{f_1}\\
U_2\ar[r]_{f_2}&X
}
\xymatrix{
U_3\bigcap U_1\ar[r]^{\iota_{3,1,1}}\ar[d]_{\iota_{3,1,3}} &U_1 \ar[d]^{f_1}\\
U_3\ar[r]_{f_3}&X
}\\
\xymatrix{
U_1\bigcap U_2\ar[r]^{\iota_{1,2,2}}\ar[d]_{\iota_{1,2,1}} &U_2 \ar[d]^{f_2}\\
U_1\ar[r]_{f_1}&X
}
\xymatrix{
U_2\ar[r]^{id_{U_2}}\ar[d]_{id_{U_2}} &U_2 \ar[d]^{f_2}\\
U_2\ar[r]_{f_2}&X
}
\xymatrix{
U_3\bigcap U_2\ar[r]^{\iota_{3,2,2}}\ar[d]_{\iota_{3,2,3}} &U_2 \ar[d]^{f_2}\\
U_3\ar[r]_{f_3}&X
}\\
\xymatrix{
U_1\bigcap U_3\ar[r]^{\iota_{1,3,3}}\ar[d]_{\iota_{1,3,1}} &U_3 \ar[d]^{f_3}\\
U_1\ar[r]_{f_1}&X
}
\xymatrix{
U_2\bigcap U_3\ar[r]^{\iota_{2,3,3}}\ar[d]_{\iota_{2,3,2}} &U_3 \ar[d]^{f_3}\\
U_2\ar[r]_{f_2}&X
}
\xymatrix{
U_3\ar[r]^{id_{U_3}}\ar[d]_{id_{U_3}} &U_3 \ar[d]^{f_3}\\
U_3\ar[r]_{f_3}&X
}
$\\
Where the inclusions $\iota_{i,j,i}=\iota_{j,i,i}, i,j=1..3$, that $\clC$ is a pre-ordered category, and $\iota_{i,j,i}$ and $\iota_{j,i,i}$ are the unique morphism $U_j\bigcap U_i=U_i\bigcap U_j\rightarrow U_i$, which we will denote by $\iota_j(i)$. Then, the above diagrams become:\\
$\xymatrix{
U_1\ar[r]^{id_{U_1}}\ar[d]_{id_{U_1}} &U_1 \ar[d]^{f_1}\\
U_1\ar[r]_{f_1}&X
}
\xymatrix{
U_2\bigcap U_1\ar[r]^{\iota_2(1)}\ar[d]_{\iota_1(2)} &U_1 \ar[d]^{f_1}\\
U_2\ar[r]_{f_2}&X
}
\xymatrix{
U_3\bigcap U_1\ar[r]^{\iota_3(1)}\ar[d]_{\iota_1(3)} &U_1 \ar[d]^{f_1}\\
U_3\ar[r]_{f_3}&X
}\\
\xymatrix{
U_1\bigcap U_2\ar[r]^{\iota_1(2)}\ar[d]_{\iota_2(1)} &U_2 \ar[d]^{f_2}\\
U_1\ar[r]_{f_1}&X
}
\xymatrix{
U_2\ar[r]^{id_{U_2}}\ar[d]_{id_{U_2}} &U_2 \ar[d]^{f_2}\\
U_2\ar[r]_{f_2}&X
}
\xymatrix{
U_3\bigcap U_2\ar[r]^{\iota_3(2)}\ar[d]_{\iota_2(3)} &U_2 \ar[d]^{f_2}\\
U_3\ar[r]_{f_3}&X
}\\
\xymatrix{
U_1\bigcap U_3\ar[r]^{\iota_1(3)}\ar[d]_{\iota_3(1)} &U_3 \ar[d]^{f_3}\\
U_1\ar[r]_{f_1}&X
}
\xymatrix{
U_2\bigcap U_3\ar[r]^{\iota_2(3)}\ar[d]_{\iota_3(2)} &U_3 \ar[d]^{f_3}\\
U_2\ar[r]_{f_2}&X
}
\xymatrix{
U_3\ar[r]^{id_{U_3}}\ar[d]_{id_{U_3}} &U_3 \ar[d]^{f_3}\\
U_3\ar[r]_{f_3}&X
}
$\\
Then one can see the obvious symmetry of rows' and columns' morphisms, those to be used to induced $\rho_0$ and $\rho_1$. Actually both diagrams that are used to induce $\rho_0$ and $\rho_1$ are identical! each of which is:\\
$\xymatrix{
&&&&\clF(U_1)\\
&&\clF(U_1)\ar[urr]^{id_{\clF(U_1)}}\ar[rr]^{\clF(\iota_2(1))}\ar[drr]_{\clF(\iota_3(1))}&&\clF(U_1\bigcap U_2)\\
&&&&\clF(U_1\bigcap U_3)\\
&&&&\clF(U_2\bigcap U_1)\\
\displaystyle\prod_{i\in \{1,2,3\}}\clF(U_i)\ar[uuurr]^{\clF(f_1)}\ar[rr]^{\clF(f_2)}\ar[dddrr]_{\clF(f_3)}&&\clF(U_1)\ar[rr]^{id_{\clF(U_2)}}\ar[urr]^{\clF(\iota_1(2))}\ar[drr]_{\clF(\iota_3(2))}&&\clF(U_2)\\
&&&&\clF(U_2\bigcap U_3)\\
&&&&\clF(U_3\bigcap U_1)\\
&&\clF(U_1)\ar[drr]_{id_{\clF(U_3)}}\ar[rr]^{\clF(\iota_2(3))}\ar[urr]^{\clF(\iota_1(3))}&&\clF(U_3\bigcap U_2)\\
&&&&\clF(U_3)
}$\\
The above argument shows that if $\clC$ is a pre-ordered category, then there is no way to get $\rho_0$ different from $\rho_1$ (up to isomorphism).\\
We notice that $\rho_0$, and $\rho_1$ are equal (up to isomorphism) basicly because of the main-diagonal-symmetry of the above matrix of fibered product diagrams. We also notice that, we cannot break this symmetry between the off-diagonal elements. Then, the trick is to try to break it in elements of the main diagonal.\\
Therefore, we work on a category $\clC$ that contains the following diagram as a fibered product diagram:\\
$\xymatrix{A\ar[r]^{\pi_0}\ar[d]_{\pi_1}&U\ar[d]^f\\
U\ar[r]_f&X}$\\
Where $\pi_0\neq \pi_1$, and $A,U,X\in \clC$\\
Then if we considered Grothendieck pre-topology on $\clC$, given by all of $2^{Ob((\clC\downarrow X))}$ as the families of coverings of every object $X$. Then for $\{F:U\rightarrow X\}$ is a covering of $X$. Then, for any functor $\clF:\clC\rightarrow \clA$, we notice that $\rho_0$ is nothing but $\clF(\pi_0)$, and $\rho_1=\clF(\pi_1)$. An example of such category $\clC$ is $\mathfrak{Set}$, take $X=\{e\}$ the set of one element, $U$ any non-empty set, then $U\times_{\{e\}}U=U\times U$, and $\pi_1,\pi_0:U\times U\rightarrow U$ s.t. $\pi_0((x,y))=x,\pi_1((x,y))=y$, or the other way around, but in both cases $\pi_0\neq \pi_1$, then the diagrams used to induce $\rho_0,\rho_1$ are different, if $\clF:\clC\rightarrow \clA$ is faithful, then $\rho_0=\clF(\pi_0)\neq  \clF(\pi_0)=\rho_0$ (Here, it is a product of one object, which is by definition the object itself).
%Is that enough, or do we need to discuss isomorphisms.
\end{proof}

